Given that sin x = 1 / sqrt(3) where x is an acute angle, determine the following: i) cos x, ii) cos2(90-x), iii) cot 2x. Starting from a right-angled triangle, determine whether sin x is a... Evaluate the following without use of figures.

Dennis, hebu tusuluhishe hili.

Tunapewa $\sin x = \frac{1}{\sqrt{5}}$ na $x$ ni pembe kali.

a) (i) Tafuta $\cos x$

Step 1: Tumia utambulisho wa Pythagorean $\sin^2 x + \cos^2 x = 1$. Badilisha thamani ya $\sin x$: $\left(\frac{1}{\sqrt{5}}\right)^2 + \cos^2 x = 1$ $\frac{1}{5} + \cos^2 x = 1$

Step 2: Tatua kwa $\cos^2 x$. $\cos^2 x = 1 - \frac{1}{5}$ $\cos^2 x = \frac{4}{5}$

Step 3: Tafuta $\cos x$. Kwa kuwa $x$ ni pembe kali, $\cos x$ itakuwa chanya. $\cos x = \sqrt{\frac{4}{5}}$ $\cos x = \frac{2}{\sqrt{5}}$ Ili kurahisisha, zidisha nambari na denominata kwa $\sqrt{5}$: $\cos x = \frac{2\sqrt{5}}{5}$ Jibu la (i) ni $\boxed{\frac{2\sqrt{5}}{5}}$.

a) (ii) Tafuta $\cos^2 (90^\circ - x)$

Step 1: Tumia utambulisho wa kikoa $\cos(90^\circ - x) = \sin x$. Hivyo, $\cos^2 (90^\circ - x) = (\sin x)^2$.

Step 2: Badilisha thamani ya $\sin x$. $\cos^2 (90^\circ - x) = \left(\frac{1}{\sqrt{5}}\right)^2$ $\cos^2 (90^\circ - x) = \frac{1}{5}$ Jibu la (ii) ni $\boxed{\frac{1}{5}}$.

a) (iii) Tafuta $\cot 2x$

Step 1: Kwanza, tafuta $\tan x$. Tunajua $\tan x = \frac{\sin x}{\cos x}$. $\tan x = \frac{\frac{1}{\sqrt{5}}}{\frac{2}{\sqrt{5}}}$ $\tan x = \frac{1}{2}$

Step 2: Tumia fomula ya $\tan 2x$: $\tan 2x = \frac{2 \tan x}{1 - \tan^2 x}$. Badilisha thamani ya $\tan x$: $\tan 2x = \frac{2 \left(\frac{1}{2}\right)}{1 - \left(\frac{1}{2}\right)^2}$ $\tan 2x = \frac{1}{1 - \frac{1}{4}}$ $\tan 2x = \frac{1}{\frac{3}{4}}$ $\tan 2x = \frac{4}{3}$

Step 3: Tafuta $\cot 2x$. Tunajua $\cot 2x = \frac{1}{\tan 2x}$. $\cot 2x = \frac{1}{\frac{4}{3}}$ $\cot 2x = \frac{3}{4}$ Jibu la (iii) ni $\boxed{\frac{3}{4}}$.

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